Q23 / 87 · NTA Abhyas (JEE Main) 2020 · Mathematics · Functions

The domain of the function $f (x) = l o g_{2} [1 - l o g_{12} (x^{2} - 5 x + 16)]$ is

Solution

The function

f (x)

is defined if

1 - l o g_{12} (x^{2} - 5 x + 16) > 0

and

x^{2} - 5 x + 16 > 0

\Rightarrow l o g_{12} (x^{2} - 5 x + 16) < 1

and

{(x - \frac{5}{2})}^{2} + \frac{39}{4} > 0

\Rightarrow x^{2} - 5 x + 16 < 12

[∵ {(x - \frac{5}{2})}^{2} + \frac{39}{4} > 0, \forall x \in R]

\Rightarrow x^{2} - 5 x + 4 < 0

\Rightarrow (x - 1) (x - 4) < 0 \Rightarrow 1 < x < 4

∴

Domain of

f = (1,4)

Hence, the correct answer is option: 'A'.